$\int \ln \left(x\right) \mathit{\ⅆ}x$

$\= x \ln \left(x\right)-\int x\cdot \left(1\/x\right) \mathit{\ⅆ}x$

$\= x \ln \left(x\right)-\int 1 \mathit{\ⅆ}x$

$\= x \ln \left(x\right)-x$

Annotated stepwise solution

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$\int \ln \left(x\right) \mathit{\ⅆ}x$$\stackrel{\text{show solution steps}}{\to }$$\begin{array}{lll}& & \text{Integration Steps}\\ & & \int \ln \left(x\right)\ⅆx\\ \text{▫}& & \text{1. Apply integration by Parts}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{Parts}}\text{rule}\\ & & \int u\ⅆv=vu-\left(\int v\ⅆu\right)\\ & \text{◦}& \text{First part}\\ & & u=\ln \left(x\right)\\ & \text{◦}& \text{Second part}\\ & & \mathrm{dv}=1\\ & \text{◦}& \text{Differentiate first part}\\ & & \mathrm{du}=\frac{\ⅆ}{\ⅆx}ln\left(x\right)\\ & & \mathrm{du}=\frac{1}{x}\\ & \text{◦}& \text{Integrate second part}\\ & & v=\int 1\ⅆx\\ & & v=x\\ & & \int \ln \left(x\right)\ⅆx=\ln \left(x\right)x-\left(\int 1\ⅆx\right)\\ & & \text{This gives:}\\ & & \ln \left(x\right)x-\left(\int 1\ⅆx\right)\\ \text{▫}& & \text{2. Apply the}\mathbf{\text{constant}}\text{rule to the term}\int 1\ⅆx\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant}}\text{rule}\\ & & \int C\ⅆx=Cx\\ & \text{◦}& \text{This means}\\ & & \int 1\ⅆx=x\\ & & \text{We can now rewrite the integral as:}\\ & & \ln \left(x\right)x-x\end{array}$

Tools≻Tasks≻Browse: Calculus - Integral≻Methods of Integration≻Parts

Table 6.1.2(a)   Integration by Parts task template